Worked Sample Problems - Math 115

For worked sample problems from Chapters 12 and 13 of the Calculus textbook, see the worked sample problems for Math 114 section of the Calculus Lab Manual.

CHAPTER 1 - Introduction to Probability

Section 1.5, page 18

Problem 8

This is inclusion/exclusion. We're given A, B, and the union U of A and B, and asked to calculate the intersection. But U=A+B-AB, so

>	AB:=A+B-U;

>	subs(A=0.5,B=0.65,U=0.85,AB);

So 30 percent of families subscribe to both newspapers.

Section 1.6, page 22

Problem 2

How many ways can the sum be even? Well, for every number on the first die, there are three numbers on the second die that add with it to make an even number -- so 3*6 = 18 times it's even out of 36 -- so the probability of an even sum is 1/2.

Section 1.7, page 27

Problem 8

Since each of the five people chooses one of the seven floors independently, there are  ways for the people to get off. But there are only 7*6*5*4*3 ways for them to get off at different floors. So the probability that this happens is

>	7!/2!/7^5;

>

evalf(%);

So the probability is only around 15%.

Section 1.8, page 34

Problem 18

The number of ways to pick students for the awards is C(100,10). But the number of ways to pick 2 students from each of the five classes is . So the probability is

>	(binomial(20,2))^5/binomial(100,10);

>

evalf(%);

So it's quite unlikely to get two from each class.

Section 1.9, page 38

Problem 6

The total number of ways for the dice to fall is .  To get every number, one of the numbers will come up twice, so you've got 6 times the multinomial coefficient (7;2,1,1,1,1,1):

>	with(combinat,multinomial);

>	6*multinomial(7,2,1,1,1,1,1)/6^7;

>

evalf(%);

So chances of getting all the numbers is a little more than 5%.

Section 1.10. page 44

Problem 6

The total number of ways to pick the 10 balls from the 90 is C(90,10). If at least one color is missing, then either all the selected balls are the same color, or they come from two colors. Call the colors R,G,B. The number of ways to pick no R's is C(60,10) (pick only from G and B), likewise for G and B. So this totals 3*C(60,10). But we've now double counted the intersections, which is the number of ways to pick only R's or only G's or only B's. This totals 3*C(30,10). There are no triple intersections, since you have to pick some color. So we have:

>	prob:=(3*binomial(60,10)-3*binomial(30,10))/binomial(90,10);

>

evalf(%);